Solve {left(x-1/x+1-x+1/x-1right)}^2=4

$${ \left( \frac{ x-1 }{ x+1 } - \frac{ x+1 }{ x-1 } \right) }^{ 2 } =4$$

Solve for x

$$x=-\left(\sqrt{2}+1\right)\approx -2.414213562$$

$$x=\sqrt{2}-1\approx 0.414213562$$

$$x=\sqrt{2}+1\approx 2.414213562$$

$$x=1-\sqrt{2}\approx -0.414213562$$

Steps Using the Quadratic Formula

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+1 and x-1 is \left(x-1\right)\left(x+1\right). Multiply \frac{x-1}{x+1} times \frac{x-1}{x-1}. Multiply \frac{x+1}{x-1} times \frac{x+1}{x+1}.

$$\left(\frac{\left(x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\right)^{2}=4$$

Since \frac{\left(x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)} and \frac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\left(\frac{\left(x-1\right)\left(x-1\right)-\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\right)^{2}=4$$

Do the multiplications in \left(x-1\right)\left(x-1\right)-\left(x+1\right)\left(x+1\right).

$$\left(\frac{x^{2}-x-x+1-x^{2}-x-x-1}{\left(x-1\right)\left(x+1\right)}\right)^{2}=4$$

Combine like terms in x^{2}-x-x+1-x^{2}-x-x-1.

$$\left(\frac{-4x}{\left(x-1\right)\left(x+1\right)}\right)^{2}=4$$

To raise \frac{-4x}{\left(x-1\right)\left(x+1\right)} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(-4x\right)^{2}}{\left(\left(x-1\right)\left(x+1\right)\right)^{2}}=4$$

Expand \left(-4x\right)^{2}.

$$\frac{\left(-4\right)^{2}x^{2}}{\left(\left(x-1\right)\left(x+1\right)\right)^{2}}=4$$

Calculate -4 to the power of 2 and get 16.

$$\frac{16x^{2}}{\left(\left(x-1\right)\left(x+1\right)\right)^{2}}=4$$

Expand \left(\left(x-1\right)\left(x+1\right)\right)^{2}.

$$\frac{16x^{2}}{\left(x-1\right)^{2}\left(x+1\right)^{2}}=4$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$\frac{16x^{2}}{\left(x^{2}-2x+1\right)\left(x+1\right)^{2}}=4$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$\frac{16x^{2}}{\left(x^{2}-2x+1\right)\left(x^{2}+2x+1\right)}=4$$

Subtract 4 from both sides.

$$\frac{16x^{2}}{\left(x^{2}-2x+1\right)\left(x^{2}+2x+1\right)}-4=0$$

Use the distributive property to multiply x^{2}-2x+1 by x^{2}+2x+1 and combine like terms.

$$\frac{16x^{2}}{x^{4}-2x^{2}+1}-4=0$$

Factor x^{4}-2x^{2}+1.

$$\frac{16x^{2}}{\left(x-1\right)^{2}\left(x+1\right)^{2}}-4=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 4 times \frac{\left(x-1\right)^{2}\left(x+1\right)^{2}}{\left(x-1\right)^{2}\left(x+1\right)^{2}}.

$$\frac{16x^{2}}{\left(x-1\right)^{2}\left(x+1\right)^{2}}-\frac{4\left(x-1\right)^{2}\left(x+1\right)^{2}}{\left(x-1\right)^{2}\left(x+1\right)^{2}}=0$$

Since \frac{16x^{2}}{\left(x-1\right)^{2}\left(x+1\right)^{2}} and \frac{4\left(x-1\right)^{2}\left(x+1\right)^{2}}{\left(x-1\right)^{2}\left(x+1\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{16x^{2}-4\left(x-1\right)^{2}\left(x+1\right)^{2}}{\left(x-1\right)^{2}\left(x+1\right)^{2}}=0$$

Do the multiplications in 16x^{2}-4\left(x-1\right)^{2}\left(x+1\right)^{2}.

$$\frac{16x^{2}-4x^{4}-8x^{3}-4x^{2}+8x^{3}+16x^{2}+8x-4x^{2}-8x-4}{\left(x-1\right)^{2}\left(x+1\right)^{2}}=0$$

Combine like terms in 16x^{2}-4x^{4}-8x^{3}-4x^{2}+8x^{3}+16x^{2}+8x-4x^{2}-8x-4.

$$\frac{24x^{2}-4x^{4}-4}{\left(x-1\right)^{2}\left(x+1\right)^{2}}=0$$

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)^{2}\left(x+1\right)^{2}.

$$24x^{2}-4x^{4}-4=0$$

Substitute t for x^{2}.

$$-4t^{2}+24t-4=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute -4 for a, 24 for b, and -4 for c in the quadratic formula.

$$t=\frac{-24±\sqrt{24^{2}-4\left(-4\right)\left(-4\right)}}{-4\times 2}$$

Do the calculations.

$$t=\frac{-24±16\sqrt{2}}{-8}$$

Solve the equation t=\frac{-24±16\sqrt{2}}{-8} when ± is plus and when ± is minus.

$$t=3-2\sqrt{2} t=2\sqrt{2}+3$$

Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for each t.

$$x=-\left(1-\sqrt{2}\right) x=1-\sqrt{2} x=\sqrt{2}+1 x=-\left(\sqrt{2}+1\right)$$